A Beautiful Equivalence of the Riemann Hypothesis

A Beautiful Equivalence of the Riemann Hypothesis José Damián Espinosa December 1, 2025 Dedication: Στην αγαπημένη μου Θεά Μαρσέλα (A mi amada Diosa Marcela) Abstract This article presents a novel, elementary and direct equivalent formulation of the famous Riemann Hypothesis. This work proposes that this hypothesis is true if and only if a specific inequality relating a simple analytical series to the sum-of-divisors function of n,σ(n) = Pd|nd, holds for all positive integers n. The equivalence, stated in the Theorem 7, allows to establish directly the hypothesis as a fine, elemental and rigorous bounding of the function σ(n)for all positive integers n. This approach offering an accessible and potential pathway to attack it. The explicit equivalence is given by: The Riemann Hypothesis is equivalent to: X k∈N 1 1+· · · +1 nk k!1 k+· · · +1 k2>X d|n d for all n∈N. “The essence of mathematics is not to make simple things complicated, but to make complicated things simple.”– S. Gudder 1 “Everything should be made as simple as possible, but not simpler.”– Albert Einstein “Truth is ever to be found in simplicity, and not in the multiplicity and confusion of things.”– Isaac Newton “Simplicity is the ultimate sophistication.”– Leonardo da Vinci “Mathematics, rightly viewed, possesses not only truth, but supreme beauty—a beauty cold and austere, like that of sculpture, without any appeal to any part of our weaker nature, without the gorgeous trappings of painting or music, yet sublimely pure, and capable of a strict perfection such as only the greatest art can show.”– Bertrand Russell “Beauty is the first test: there is no permanent place in the world for ugly mathematics.”– G. H. Hardy “A mathematician is not complete until he is a bit of a poet in his soul.”– Sofia Kovalevskaya “The scientist does not study nature because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.”– Henri Poincaré “Mathematics possesses not only truth, but supreme beauty.”– Bertrand Russell “Imagination is more important than knowledge. Knowledge is limited. Imagination encircles the world.”– Albert Einstein 2 “Logic will get you from A to B. Imagination will take you everywhere.”– Albert Einstein “If I had an hour to solve a problem, I’d spend 55 minutes thinking about the problem and 5 minutes thinking about solutions.”– Albert Einstein “An expert is a person who has made all the mistakes that can be made in a very narrow field.”– Niels Bohr “To achieve the impossible, you must attempt the absurd.”– Miguel de Cervantes “We must know, we will know (Wir müssen wissen. Wir werden wissen.).”– David Hilbert Contents 1 Introduction 4 2 Content 5 2.1 Novel Equivalence of the Riemann Hypothesis . . . . . . . . . . . 5 2.1.1 Preliminary Concepts . . . . . . . . . . . . . . . . . . . . 5 2.1.2 About Pk∈N(Hk−γ)xk k!................... 8 2.1.3 Elementary Versions of the Riemann Hypothesis by Lagarias (2002) and Robin (1984) . . . . . . . . . . . . . . . 10 2.1.4 About K(x):=Pk∈N xk k!M(k)................ 13 2.1.5 Espinosa’s Riemann Hypothesis Equivalence (2025) . . . . 28 References 39 3 1 Introduction The Riemann Hypothesis, first proposed by Bernhard Riemann in 1859, stands as one of the most profound and challenging unsolved problems in modern mathematics. It asserts that all non-trivial zeros of the Riemann zeta function ζ(s)lie on the critical line Re(s) = 1 2. The veracity of this, has deep and far-reaching implications, extending into the core of analytic number theory, particularly in determining the precise distribution of the prime numbers. Its resolution would necessitate a fundamental re-evaluation of numerous established results conditional upon its truth. Due to the immense difficulty in proving it in its original form, a significant body of research has been dedicated to establishing equivalent formulations. This approach translates the problem from the complex plane into elementary number theory or analysis, offering new avenues for attack. A prominent line of inquiry involves connecting the hypothesis to inequalities governing the sumof-divisors function, σ(n) = Pd|nd. For instance, in 1984, Robin [Rob84] demonstrated that the Riemann Hypothesis is equivalent to the statement that the inequality σ(n)< eγnlog log n holds for all integers n≥5041, where γis the Euler-Mascheroni constant. Similarly, a distinct but equally powerful formulation was provided by Lagarias [Lag02] in 2002, who proved that the Riemann Hypothesis holds if and only if σ(n)≤Hn+eHnlog(Hn)for all n∈N, where Hn=Pn i=1 1 iis the n-th harmonic number. These equivalences highlight the deep connection between the distribution of primes and fundamental arithmetic functions. Following this tradition of analytical reformulation, the present article introduces a novel and powerful equivalence for the Riemann Hypothesis. This new formulation is cast as an elementary inequality that relates the sum-ofdivisors function, σ(n), to a elemental analytical expression series. The equivalence is stated in the Theorem 7, and allows to establish directly the hypothesis as a fine, elemental and rigorous bounding of the function σ(n)for all positive integers n. This approach offering an accessible, powerful and potential pathway to attack the hypothesis. Our main result is the proof that the Riemann Hypothesis is true if and only if the following inequality holds for all positive integers n∈N: X k∈N1 1+· · · +1 nk k!1 k+···+1 k2>X d|n d The remainder of this paper is structured as follows. Section 2 (Content) provides the complete development of this work. Within this, Section 2.1 (Novel Equivalence of the Riemann Hypothesis) meticulously develops the analytical framework for our main result. We begin in Section 2.1.1 (Preliminary Concepts) by establishing the necessary concepts, definitions and properties used in our formulation. This is followed by an exploration of a related series in Section 2.1.2 (About Pk∈N(Hk−γ)xk k!), we analyze the convergence, some properties and equalities 4 related, and a specific asymptotic expansion of the series Pk∈N(Hk−γ)xk k!, also we establish upper and lower bounds for its value. Then we present the Section 2.1.3 (Elementary Versions of the Riemann Hypothesis by Lagarias (2002) and Robin (1984)) briefly reviews the established inequality equivalences upon which our work builds, connecting the hypothesis to elementary number theory. Immediately following, we introduce and define the function K(x)in Section 2.1.4 (About K(x):=Pk∈N xk k!M(k)). Here, we construct and prove its convergence, study its properties, derive its asymptotic expansion, and establish upper and lower bounds for K(x)and M(k)respectively. This is the series that is the centerpiece of the new equivalence of the Riemann Hypothesis, which we use in the next section, where the continuous development of these analytical results culminates. Ultimately in the final Section 2.1.5 (Espinosa’s Riemann Hypothesis Equivalence (2025)), we state and rigorously prove Theorem 7, the proposed new equivalent version of the Riemann Hypothesis, establishing the validity of our novel equivalence. We connect the K(x):=Pk∈N xk k!M(k), using results from sections 2.1.1, 2.1.2, and 2.1.4, with the known equivalences by Lagarias and Robin (Section 2.1.3) resulting in establish and prove the equivalence of our new inequality to the Riemann Hypothesis. Crucially, the elementary and beauty structure of the proposed inequality, which leverages properties, represents a significant improvement in elementarity compared to traditional approaches derived from the complex analysis of the zeta function. This formulation opens a new analytical avenue for attacking the problem within the realm of elementary number theory and combinatorics. 2 Content 2.1 Novel Equivalence of the Riemann Hypothesis In this Section 2.1, we work with a continuous interweaving of results until reaching Theorem 7, the proposed new version of the Riemann Hypothesis. This set is presented through ordered sections that build upon one another as the narrative progresses, constructively, following the clear natural line that finally leads to the long-awaited aforementioned result. 2.1.1 Preliminary Concepts This Section 2.1.1 addresses key concepts, which lay the foundation for everything undertaken, for what is constructed later. It begins by defining the famous Euler-Mascheroni Constant γ, an essential constant when addressing the Riemann Hypothesis, the definition follows: Definition 1 (Euler-Mascheroni Constant γ).The Euler-Mascheroni constant 5 γis defined by the limit (Abramowitz [AS64, Formula 6.1.3]): γ:= lim n→∞   n X k=1 1 k−log n = 0.5772156649 ... This limit exists and is finite, representing the asymptotic difference between the harmonic series and the natural logarithm. Having established these results, we proceed to mention and define some fundamental concepts, such as convergence and radius of convergence of real series as follows in Definition 2: Definition 2 (Convergence of real power series).Let P∞ k=0 ckxkbe a power series with ck, x ∈R. We say that (Rudin [Rud64, Definition 3.38]): •The series converges at x∈Rif limn→∞ Pn k=0 ckxkexists and is finite. •The radius of convergence R∈[0,+∞]is the supremum of the |x|for which the series converges. Then we state the Series Comparison Test, of enormous utility in Theorem 1: Theorem 1 (Comparison Test for Series).Let {ak}k∈N,{bk}k∈N⊂R+ 0be sequences of non-negative real numbers such that: 0≤bk≤ak∀k∈N. If the series P∞ k=1 akconverges, then P∞ k=1 bkalso converges. Proof. See p. 60 in Rudin [Rud64, Theorem 3.25]). Next, we establish the Monotone Convergence Theorem for Series, also extremely useful in Theorem 2: Theorem 2 (Monotone Convergence Theorem for Series).A series P∞ k=1 ak with non-negative terms, i.e., ak≥0for all k∈N, converges if and only if the sequence of its partial sums SN=PN k=1 akis bounded above. That is, there exists M∈Rsuch that SN≤Mfor all N∈N. And two other important results of the series in these two propositions stated immediately, in Proposition 1 and 2, respectively: Proposition 1. Let {an}n∈Nbe a sequence of real numbers that satisfies the following two conditions: 1. The sequence is decreasing, i.e., an+1 ≤anfor all n∈N. 2. The limit of the sequence is zero, i.e., limn→∞ an= 0. Then, all terms of the sequence are non-negative, i.e., an≥0for all n∈N. 6 Proposition 2. Let {an}n∈Nbe a sequence of real numbers that satisfies the following two conditions: 1. The sequence is increasing, i.e., an+1 ≥anfor all n∈N. 2. The limit of the sequence is zero, i.e., limn→∞ an= 0. Then, all terms of the sequence are non-positive, i.e., an≤0for all n∈N. Similarly, the concept of Asymptotic Upper Bound for a real function is defined through Definition 3: Definition 3 (ONotation (Asymptotic Upper Bound)).Let f, g:N→R(or f, g :R→R). We say that: f(x) = O(g(x)) as x→ ∞, if there exist constants x0>0and M > 0such that: |f(x)| ≤ M|g(x)|for all x≥x0. And its fundamental properties as follows in Proposition 3: Proposition 3 (Fundamental Properties of the ONotation).Let E⊆Rand let f1, f2, g1, g2:E→Rbe functions. For k∈R\ {0}, the following hold: (T) Transitivity:f1=O(g1)∧g1=O(g2) =⇒f1=O(g2); (M) Multiplication by Function (set equality): f2· O(g1)=O(f2g1); (S) Sum:f1=O(g1)∧f2=O(g2) =⇒f1+f2=O(max(|g1|,|g2|)); (H) Homogeneity:f1=O(g1) =⇒kf1=O(g1). Proposition 4 (Bounds for the Exponential Integral).For x∈R+, with x= 0, the function E1(x)satisfies the following inequalities: 1 2log 1 + 2 x< exE1(x)<log 1 + 1 x, where: •E1(x) = R∞ x e−t tdt is the first-order exponential integral. Proof. See p. 229 in Abramowitz [AS64, Formula 5.1.20]. 7 2.1.2 About Pk∈N(Hk−γ)xk k! Given all the definitions in Section 2.1.1, we begin with a series, a very special one, of central importance in this work, as will be seen later. The advantage offered by Proposition 5 is phenomenal: it allows writing the closed expression exlog xas a simple power series involving the function ψ(k+1), plus a very small error of size O1 x, instead of requiring the product of the corresponding Taylor or other asymptotic series for exand log x, respectively. It is stated below: Proposition 5. Let x∈R+, the following holds: X k∈N0 ψ(k+ 1)xk k!=exlog x+O1 xas x→ ∞, where: •ψ(k)=Γ′(k)/Γ(k)is the digamma function; •Γ(k)is the gamma function; •N0={0,1,2, . . .}denotes the set of natural numbers including zero. Proof. See pp. 31-33 in Dingle [Din73]. Then we proceed to give its exact expression in Proposition 6 below: Proposition 6. For x∈R+, the following holds: X k∈N0 ψ(k+ 1)xk k!=exlog x+E1(x), where: •ψ(k)=Γ′(k)/Γ(k)is the digamma function; •Γ(k)is the gamma function; •N0={0,1,2, . . .}denotes the set of natural numbers including zero; •E1(x) = R∞ xu−1e−udu is the exponential integral. Proof. See pp. 31-33 in Dingle [Din73]. Next, in Proposition 7, we have a highly useful relationship between ψ(z) and the n-th harmonic number Hn: Proposition 7. Let n∈Z≥1, the digamma function satisfies: ψ(n) = −γ+Hn−1 where: 8 •Hn=Pn i=1 1 iis the n-th harmonic number (H0:= 0); •γis the Euler-Mascheroni constant. Proof. See p. 258 in Abramowitz [AS64, Formula 6.3.1]. We therefore proceed to modify the series in Proposition 5 and adapt the result derived from it, expressing it no longer in terms of the digamma function ψ(k), which is a complicated value, but in terms of the harmonic number Hk. All this is seen in Proposition 8 which is stated next: Proposition 8. Let x∈R+, the following holds: X k∈N (Hk−γ)xk k!=exlog x+γ+O1 xas x→ ∞, where: •N={1,2,3, . . .}denotes the set of natural numbers; •Hk=Pk n=1 1 nis the k-th harmonic number (H0:= 0); •γis the Euler-Mascheroni constant. Proof. We start from Proposition 5: ∞ X k=0 ψ(k+ 1)xk k!=exlog x+O1 x.(1) We apply Proposition 7 which relates the digamma function to harmonic numbers: ψ(k+ 1) = Hk−γ, ∀k≥0.(2) Substituting (2) into (1): ∞ X k=0 (Hk−γ)xk k!=exlog x+O1 x. We explicitly separate the term k= 0: (H0−γ) + ∞ X k=1 (Hk−γ)xk k!=exlog x+O1 x. We evaluate H0(by definition H0= 0): (0 −γ) + ∞ X k=1 (Hk−γ)xk k!=exlog x+O1 x. Passing −γto the other side: ∞ X k=1 (Hk−γ)xk k!=exlog x+γ+O1 x. 9 Proof. The objective is to prove that the left-hand side inequality, Hk−γ < M(k), is true for every natural number k. We can rewrite this inequality as: M(k)−(Hk−γ)>0. Substituting the definition M(k) = Pk2 i=k1 i(which can also be expressed as Hk2−Hk−1) and rearranging the terms, the inequality becomes equivalent to demonstrating that the following expression is always greater than zero: δk=Hk2−(Hk−1+Hk)+γ > 0. The demonstration of this result is based on two key arguments that, when combined, prove that all terms of the sequence δkare positive. Demonstration of the Limit of the Sequence For this argument, we use the asymptotic formula for the n-th harmonic number, which is derived from the Euler-Maclaurin formula, given by the expansion of Theorem 6: Hn= log n+γ+1 2n+O1 n2. Substituting this formula into the expression for δk, we obtain: δk= log(k2)+γ+1 2k2+O1 k4! − log(k) + γ+1 2k+O1 k2! − log(k−1)+γ+1 2(k−1) +O1 (k−1)2! +γ. Now, we explicitly group the terms by type: δk=log(k2)−log(k)−log(k−1) + (γ−γ−γ+γ) +1 2k2−1 2k−1 2(k−1) + O1 k4− O 1 k2− O 1 (k−1)2!. We observe the following: •The constant terms γcancel directly: γ−γ−γ+γ= 0. 16 •For the Oterms, we note that O1 (k−1)2=O1 k2. When summing and subtracting terms of different orders, the term of the lowest order dominates. In this case, the term of the lowest order is O1 k2. Therefore, the complete sum of the Oterms is O1 k2. •The algebraic terms 1 2k2−1 2k−1 2(k−1) are of order O1 k, since 1 kis the term of greatest magnitude. Thus, the simplified expression is: δk=log(k2)−log(k)−log(k−1)+O1 k. Using the properties of logarithms, the first parenthesis becomes: log(k2)−log(k)−log(k−1) = log k2 k(k−1)!= log k k−1. Now, we take the limit of the complete expression: lim k→∞ δk= lim k→∞ "log k k−1+O1 k#. We evaluate each term of the limit separately: lim k→∞ log k k−1= log lim k→∞ k k−1= log(1) = 0 lim k→∞ O1 k= 0.(by definition of the Onotation) Adding both limits, the final result is: lim k→∞ δk= 0 + 0 = 0. This rigorously demonstrates that the sequence of terms δkconverges to zero. Demonstration that the Sequence is Decreasing To demonstrate that the sequence is decreasing, we must prove that the difference between consecutive terms is negative, i.e., δk+1 −δk<0. The complete calculation of this difference is: δk+1 −δk=H(k+1)2−Hk−Hk+1 +γ −(Hk2−Hk−1−Hk+γ). 17 Substituting the harmonic series identities (H(k+1)2−Hk2=P(k+1)2 j=k2+1 1 j,Hk− Hk−1=1 k,Hk+1 −Hk=1 k+1 ), the expression simplifies to the complete calculation of this difference: δk+1 −δk=H(k+1)2−Hk2−(Hk−Hk−1)−(Hk+1 −Hk) = (k+1)2 X j=k2+1 1 j−1 k−1 k+ 1. Rewriting the sum with a change of index, j=i+ (k2+k+ 1): (k+1)2 X j=k2+1 1 j= k X i=−k 1 (k2+k+ 1) + i. Using the property of symmetric series (Pk i=−k1 x+i=1 x+Pk i=1 1 x+i+1 x−i), the expression becomes: δk+1 −δk=1 k2+k+ 1 + k X i=1 1 (k2+k+ 1) + i+1 (k2+k+1)−i−2k+ 1 k(k+ 1) =1 k2+k+ 1 + k X i=1 2(k2+k+ 1) (k2+k+ 1)2−i2−2k+ 1 k2+k. Now, we expand the term inside the sum using 2x x2−j2=2 x+2j2 x(x2−j2)with x=k2+k+ 1: δk+1 −δk=1 k2+k+ 1 + k X i=1 2 k2+k+ 1 +2i2 (k2+k+ 1)((k2+k+ 1)2−i2)! −2k+ 1 k2+k =1 k2+k+ 1 +2k k2+k+ 1 + k X i=1 2i2 (k2+k+ 1)((k2+k+ 1)2−i2)! −2k+ 1 k2+k =2k+ 1 k2+k+ 1 + k X i=1 2i2 (k2+k+ 1)((k2+k+ 1)2−i2)!−2k+ 1 k2+k =2k+ 1 k2+k+ 1 + 2 k X j=1 j2 (k2+k+ 1)((k2+k+ 1)2−j2)−2k+ 1 k2+k. Factoring the term 1 k2+k+1 and reorganizing, we obtain: δk+1−δk=1 k2+k+ 1  2k+1+ k X j=1 2j2 (k2+k+ 1)2−j2−2k+ 1 k2+k(k2+k+ 1) . 18 Simplifying the term −2k+1 k2+k(k2+k+ 1) and rearranging, we obtain the final expression: δk+1−δk=1 k2+k+ 1   k X j=1 2j2 (k2+k+ 1)2−j2−2k+ 1 k2+k(k2+k+ 1) + 2k+ 1 . We calculate the combined term: −2k+ 1 k2+k(k2+k+ 1) + (2k+ 1) = (2k+ 1) 1−k2+k+ 1 k2+k! = (2k+ 1) k2+k−(k2+k+ 1) k2+k!= (2k+ 1) −1 k2+k=−2k+ 1 k2+k. Therefore: δk+1 −δk=1 k2+k+ 1  −2k+ 1 k2+k+ k X j=1 2j2 (k2+k+ 1)2−j2 . To prove that δk+1 −δk<0, it is sufficient to demonstrate that the term inside the parenthesis is negative. This reduces to demonstrating that: 2k+ 1 k2+k> k X j=1 2j2 (k2+k+ 1)2−j2. To prove this inequality, we bound the sum on the right-hand side. The denominator (k2+k+ 1)2−j2is always greater than (k2+k+ 1)2−k2= (k2+k+ 1 −k)(k2+k+1+k)=(k2+ 1)(k2+ 2k+ 1) >(k2+k)(k2+k+ 1). Therefore, we can establish the following inequality: 1 (k2+k+ 1)2−j2<1 (k2+k)(k2+k+ 1). Multiplying by 2j2≥0both sides of the inequality, we obtain: 2j2 (k2+k+ 1)2−j2<2j2 (k2+k)(k2+k+ 1). Summing for each jwe have: k X j=1 2j2 (k2+k+ 1)2−j2< k X j=1 2j2 (k2+k)(k2+k+ 1) =2 (k2+k)(k2+k+ 1) k X j=1 j2. We know that Pk j=1 j2=k(k+1)(2k+1) 6. Substituting this into the inequality, we obtain: 2 (k2+k)(k2+k+ 1) k(k+ 1)(2k+ 1) 6=2k+ 1 3(k2+k+ 1) 19 k X j=1 2j2 (k2+k+ 1)2−j2<2k+ 1 3(k2+k+ 1). The inequality we need to prove is then: 2k+ 1 k2+k>2k+ 1 3(k2+k+ 1). Demonstration of the Final Left-Hand Side Inequality To prove that 2k+1 k2+k>2k+1 3(k2+k+1) , since k≥1, the term 2k+ 1 is positive, so we can divide by it without changing the direction of the inequality: 1 k2+k>1 3(k2+k+ 1). Taking the reciprocal of both sides (which reverses the inequality): k2+k < 3(k2+k+ 1). Distributing the 3 on the right-hand side: k2+k < 3k2+ 3k+ 3. Subtracting the terms from the left-hand side from both sides: 0<2k2+ 2k+ 3. For all k≥1,2k2is positive, 2kis positive, and 3is positive. The sum of three positive terms is always positive. Therefore, the inequality 0<2k2+ 2k+ 3 is true for all k≥1. This proves that δk+1 −δk<0for all k∈N, so the sequence is strictly decreasing. We have demonstrated that the sequence δkconverges to zero and is strictly decreasing. By Proposition 1, a sequence with these two properties must be composed of non-negative terms. Therefore, δk≥0for all k∈N, which proves the original left-hand side inequality. Now the objective is to prove that the right-hand side inequality, M(k)< Hk−γ+1 k, is true for every natural number k. We can rewrite this inequality as: M(k)−Hk−γ+1 k<0. Substituting the definition M(k) = Pk2 i=k1 i(which can also be expressed as Hk2−Hk−1) and rearranging the terms, the inequality becomes equivalent to demonstrating that the following expression is always less than zero: δk=Hk2−Hk−1−Hk+γ−1 k<0. 20 The demonstration of this result is based on two key arguments that, when combined, prove that all terms of the sequence δkare negative. Demonstration of the Limit of the Sequence For this argument, we use the asymptotic formula for the n-th harmonic number, which is derived from the Euler-Maclaurin formula: Hn= log n+γ+1 2n+O1 n2. Substituting this formula into the expression for δk, we obtain: δk= log(k2)+γ+1 2k2+O1 k4! − log(k) + γ+1 2k+O1 k2! − log(k−1)+γ+1 2(k−1) +O1 (k−1)2! +γ−1 k. Now, we explicitly group the terms by type: δk=log(k2)−log(k)−log(k−1) + (γ−γ−γ+γ) +1 2k2−1 2k−1 2(k−1) −1 k + O1 k4− O 1 k2− O 1 (k−1)2!. We observe the following: •The constant terms γcancel directly: γ−γ−γ+γ= 0. •For the Oterms, we note that O1 (k−1)2=O1 k2. When summing and subtracting terms of different orders, the term of the lowest order dominates. In this case, the term of the lowest order is O1 k2. Therefore, the complete sum of the Oterms is O1 k2. •The algebraic terms 1 2k2−1 2k−1 2(k−1) −1 kare of order O1 k, since 1 k is the term of greatest magnitude. Thus, the simplified expression is: δk=log(k2)−log(k)−log(k−1)+O1 k. 21 Using the properties of logarithms, the first parenthesis becomes: log(k2)−log(k)−log(k−1) = log k2 k(k−1)!= log k k−1. Now, we take the limit of the complete expression: lim k→∞ δk= lim k→∞ "log k k−1+O1 k#. We evaluate each term of the limit separately: lim k→∞ log k k−1= log lim k→∞ k k−1= log(1) = 0 lim k→∞ O1 k= 0.(by definition of the Onotation) Adding both limits, the final result is: lim k→∞ δk= 0 + 0 = 0. This rigorously demonstrates that the sequence of terms δkconverges to zero. Demonstration that the Sequence is Increasing To demonstrate that the sequence is increasing, we must prove that the difference between consecutive terms is positive, i.e., δk+1 −δk>0. The complete calculation of this difference is: δk+1 −δk=H(k+1)2−Hk−Hk+1 +γ−1 k+ 1 −Hk2−Hk−1−Hk+γ−1 k. Substituting the harmonic series identities (H(k+1)2−Hk2=P(k+1)2 j=k2+1 1 j,Hk− Hk−1=1 k,Hk+1 −Hk=1 k+1 ), the expression simplifies to: δk+1 −δk=   (k+1)2 X j=k2+1 1 j  −1 k−1 k+ 1 −1 k+ 1 +1 k = (k+1)2 X j=k2+1 1 j−2 k+ 1. To prove that δk+1 −δk>0, we need to prove that P(k+1)2 j=k2+1 1 j>2 k+1 . 22 Now, we will expand the expression P(k+1)2 j=k2+1 1 janalogously to the original. We rewrite the sum with a change of index, j=i+ (k2+k+ 1): (k+1)2 X j=k2+1 1 j= k X i=−k 1 (k2+k+ 1) + i. Using the property of symmetric series (Pk i=−k1 x+i=1 x+Pk i=1 1 x+i+1 x−i), the expression becomes: k X i=−k 1 (k2+k+ 1) + i=1 k2+k+ 1 + k X i=1 1 (k2+k+ 1) + i+1 (k2+k+1)−i. We apply the identity 1 x+j+1 x−j=2x x2−j2: =1 k2+k+ 1 + k X i=1 2(k2+k+ 1) (k2+k+ 1)2−i2. Now, we expand the term inside the sum using 2x x2−j2=2 x+2j2 x(x2−j2)with x=k2+k+ 1: =1 k2+k+ 1 + k X i=1 2 k2+k+ 1 +2i2 (k2+k+ 1)((k2+k+ 1)2−i2)! =1 k2+k+ 1 +2k k2+k+ 1 + k X i=1 2i2 (k2+k+ 1)((k2+k+ 1)2−i2) =2k+ 1 k2+k+ 1 + 2 k X j=1 j2 (k2+k+ 1)((k2+k+ 1)2−j2). Thus, the complete difference δk+1 −δkis: δk+1 −δk=  2k+ 1 k2+k+ 1 + 2 k X j=1 j2 (k2+k+ 1)((k2+k+ 1)2−j2) −2 k+ 1. Now, we combine the terms that are not in the sum: 2k+ 1 k2+k+ 1 −2 k+ 1 =(2k+ 1)(k+1)−2(k2+k+ 1) (k2+k+ 1)(k+ 1) =2k2+ 3k+1−(2k2+ 2k+ 2) (k2+k+ 1)(k+ 1) =k−1 (k2+k+ 1)(k+ 1). 23 Therefore, the final expression for the difference δk+1 −δkis: δk+1 −δk=k−1 (k2+k+ 1)(k+ 1) + 2 k X j=1 j2 (k2+k+ 1)((k2+k+ 1)2−j2). Demonstration of the Final Right-Hand Side Inequality Now we prove that δk+1 −δk>0for all k≥1: •The first term, k−1 (k2+k+1)(k+1) , is non-negative for all k≥1. This is because the numerator (k−1) is equal to 0when k= 1, and is positive for any k > 1. The denominator (k2+k+ 1)(k+ 1) is always positive for any natural number k≥1. •The second term, 2Pk j=1 j2 (k2+k+1)((k2+k+1)2−j2), is a sum of strictly positive terms for all k≥1. For any jin the range of the sum (from 1to k), the numerator j2is strictly positive. The denominator (k2+k+ 1)((k2+ k+ 1)2−j2)is also strictly positive, since j2≤k2, and k2is strictly less than (k2+k+ 1)2, ensuring that the term (k2+k+ 1)2−j2is positive. Therefore, each individual term of this sum is strictly positive. The sum of a non-negative term and a strictly positive term always results in a strictly positive value. Therefore, δk+1 −δk>0for all k≥1. This proves that the sequence δkis strictly increasing. We have demonstrated that the sequence δkconverges to zero and is strictly increasing. By Proposition 2, a sequence that is strictly increasing and whose limit is zero must be composed of non-positive terms. Therefore, δk≤0for all k∈N, which proves the original right-hand side inequality. Finally, we have reached the result we wanted to demonstrate; we have the sought-after bounding, both left and right respectively, of the function M(k). Next, we proceed to state the two key results for bounding and controlling the behavior of the function K, in Proposition 12 and 13 respectively: Proposition 12. For x∈R+, it holds that K(x)> exlog x. Proof. Let us fix an arbitrary x>0. Recall the definition of the function K(x)from Definition 4: K(x):=X k∈N xk k!M(k). By Proposition 11, we know that for every integer k≥1: Hk−γ < M(k). Multiplying both sides of the inequality Hk−γ < M(k)by xk k!(which is a positive term, since x>0), we obtain: (Hk−γ)xk k!<M(k)xk k!. 24 Summing over all values of k∈N(from k= 1 to ∞), we establish the following relationship for the series: X k∈N (Hk−γ)xk k!<X k∈N M(k)xk k!. By Definition 4, the right-hand side of this inequality is precisely K(x): X k∈N (Hk−γ)xk k!<K(x). Now, let’s consider the left part of this inequality. By Proposition 9, we know that for any x∈R+: exlog x+1 2log 1 + 2 x+γ < X k∈N (Hk−γ)xk k!. Combining the two inequalities obtained, we have the following chain: exlog x+1 2log 1 + 2 x+γ < X k∈N (Hk−γ)xk k!<K(x). From this chain of inequalities, we can directly infer that: exlog x+1 2log 1 + 2 x+γ < K(x). Since x∈R+, the terms 1 2log 1 + 2 xand γare both positive. Specifically, log 1 + 2 x>0because 1 + 2 x>1, and the Euler-Mascheroni constant γ≈ 0.57721 is positive. Therefore, the sum of these two terms, 1 2log 1 + 2 x+γ, is strictly greater than zero. This implies that: exlog x < exlog x+ 1 2log 1 + 2 x+γ!. Finally, by combining this last inequality with the one we previously established: exlog x < exlog x+1 2log 1 + 2 x+γ!<K(x), we conclude that: exlog x<K(x). This demonstration is valid for all x∈R+. Proposition 13. For all x∈R+, it holds that: K(x) = exlog x+Oex xas x→ ∞. 25 Figure 1: Bound of Espinosa’s Inequality from Theorem 7 for the function σ(n), for values of n∈Nwith n≤100. Figure 2: Bound of Espinosa’s Inequality from Theorem 7 for the function σ(n), for values of n∈Nwith n≤1000. 32 Figure 3: Bound of Espinosa’s Inequality from Theorem 7 for the function σ(n), for values of n∈Nwith n≤10000. Figure 4: Coefficient akof the Bound of Espinosa’s Inequality from Theorem 7 for the function σ(n), for values of k∈Nwith k≤100. 33 Figure 5: Coefficient akof the Bound of Espinosa’s Inequality from Theorem 7 for the function σ(n), for values of k∈Nwith k≤1000. A more compact form is proposed below. Corollary 2 (Espinosa, 2025).The Riemann Hypothesis is equivalent to: X k∈N Hk n k!ak> σ(n) for all n∈N, where: •Hn=1 1+· · · +1 n, •ak=1 k+· · · +1 k2, •σ(n) = Pd|ndis the sum-of-divisors function of n. Next, the most reduce form of this version of the infamous Riemann Hypothesis is stated. Corollary 3 (Espinosa, 2025).The Riemann Hypothesis is equivalent to: X k∈N1 1+· · · +1 nk k!1 k+···+1 k2>X d|n d for all n∈N. 34 Figure 6: Bound of Espinosa’s Inequality from Theorem 7 for the function σ(n), for values of n∈Nwith n≤1000. Figure 7: Bound of Espinosa’s Inequality from Theorem 7 for the function σ(n), for values of n∈Nwith n≤1000. 35 Figure 8: Bound of Espinosa’s Inequality from Theorem 7 for the function σ(n), for values of n∈Nwith n≤10000. Figure 9: Bound of Espinosa’s Inequality from Theorem 7 for the function σ(n), for values of n∈Nwith n≤10000. 36 Figure 10: Bound of Espinosa’s Inequality from Theorem 7 for the function σ(n), for values of n∈Nwith n≤10000. Figure 11: Bound of Espinosa’s Inequality from Theorem 7 for the function σ(n), for values of n∈Nwith n≤10000. 37 Figure 12: Bound of Espinosa’s Inequality from Theorem 7 for the function σ(n), for values of n∈Nwith n≤100000. Figure 13: Bound of Espinosa’s Inequality from Theorem 7 for the function σ(n), for values of n∈Nwith n≤100000. 38 References [Apo76] T. M. Apostol. Introduction to Analytic Number Theory. Teorema 3.2 (pp. 55-58). Springer, 1976. url:https://dl.icdst.org/pdfs/fil es1/ebc2974176a03ab93756026a97b6d370.pdf. [AS64] M. Abramowitz and I. A. Stegun. Handbook of Mathematical Functions. Fórmula 6.3.1 (p. 258). Fórmula 6.1.3 (p. 255). Dover, 1964. url:https://archive.org/details/handbookofmathem1964abra. [Bru81] N. G. de Bruijn. Asymptotic Methods in Analysis. Sección 1.2 (pp. 48). Definición rigurosa: Ec. 1.2.3 (p. 5). Propiedades de transitividad: Teorema 1.2 (p. 7). Dover, 1981. url:https://store.doverpublic ations.com/0486642216.html. [Cor09] T. H. Cormen. Introduction to Algorithms. 3rd ed. Capítulo 3 (Growth of Functions), Sección 3.1 (pp. 43-50). Definición formal de O-grande: Teorema 3.1 (p. 45). Propiedades algebraicas: Lema 3.1 (p. 47). MIT Press, 2009. url:https://mitpress.mit.edu/books/introductio n-algorithms. [Din73] R. B. Dingle. Asymptotic Expansions: Their Derivation and Interpretation. London: Academic Press, 1973. url:https://michaelberry physics.wordpress.com/wp-content/uploads/2013/07/dingle.p df. [Lag02] J. C. Lagarias. “An Elementary Problem Equivalent to the Riemann Hypothesis”. In: American Mathematical Monthly (2002). Teorema 1.1 (p. 1). Teorema 3.1 (p. 6). Teorema 3.2 (p. 7). Conclusión (p. 10). url:https://dept.math.lsa.umich.edu/~lagarias/doc/elemen taryrh.pdf. [Rob84] G. Robin. “Grandes valeurs de la fonction somme des diviseurs et hypothèse de Riemann”. In: Journal de Mathématiques Pures et Appliquées 63.2 (1984), pp. 187–213. url:https://www.sciencedirec t.com/science/article/pii/0021782484901676. [Rud64] W. Rudin. Principles of Mathematical Analysis. 3rd ed. Teorema 3.25 (p. 60). Definición 3.38 y Teorema 3.39 (p. 69). McGraw-Hill, 1964. url:https://david92jackson.neocities.org/images/Principl es_of_Mathematical_Analysis-Rudin.pdf. 39